If $A^T = -A$ for an $n \times n$ matrix $A$ with zeros on the diagonal, is the rank at most $n-1$? I was looking at a textbook problem to compute the rank of a matrix $A$ with $A^T = -A$ and zeros on the diagonal, and I do not know how to approach the problem other than going through elimination steps involving relatively tedious arithmetic. For example, if $n = 5$, the matrix might be as below. The answer for a $5 \times 5$ matrix I know is at most 4 (less possibly). Does anyone know a neat way to show this and is it true for any $n$?
$$ \left( \begin{array}{ccccc}
0 & a & b & c & d \\
-a & 0 & e & f & g \\
-b & -e & 0 & h & k \\
-c & -f & -h & 0 & l \\
-d & -g & -k & -l & 0
 \end{array} \right) $$
 A: You are asking if the rank of an antisymmetric matrix is at most $n-1$.
If $n$ is odd, you have that:
$$\det(A)=\det(A^T)=\det(-A)=(-1)^n \det(A)=-\det(A)$$
Thus, $\det(A)=0$, so the rank must be less than $n$, else it would have non-zero determinant.
If $n$ is even, no such constraint exists, and the rank can be $n$. e.g.:
$$\operatorname{rank}\left( \begin{array}{}
   0 & 1 \\
   -1 & 0 \\
  \end{array}  \right)=2$$
A: An example of a $2 \times 2$ antisymmetric matrix with rank $2$ is $\pmatrix{0 & 1\cr -1 & 0\cr}$.  String these along the diagonal and you get for any even $n$ an antisymmetric matrix with rank $n$.
EDIT: For extra credit: the rank of an antisymmetric matrix (over the real numbers) is always even.  This is because the nonzero eigenvalues are imaginary, and thus come in complex-conjugate pairs.
A: since, it is $5 \times 5$ determinant must be zero. consider the the sub matrix obtained by eliminating 5 th column and 5 th row. We get $B$(say). B is also anti symmetric. but order is $4\times4$. If all the eigen values are non zero then determinant is non zero. Rank is 4.
